A076725 a(n) = a(n-1)^2 + a(n-2)^4, a(0) = a(1) = 1.

1, 1, 2, 5, 41, 2306, 8143397, 94592167328105, 13345346031444632841427643906, 258159204435047592104207508169153297050209383336364487461

Offset: 0

Author

Michael Somos, Oct 29 2002

Keywords

• nonn
• nice

Comments

a(n) and a(n+1) are relatively prime for n >= 0.

The number of independent sets on a complete binary tree with 2^(n-1)-1 nodes. - Jonathan S. Braunhut (jonbraunhut(AT)usa.net), May 04 2004. For example, when n=3, the complete binary tree with 2 levels has 2^2-1 nodes and has 5 independent sets so a(3)=5. The recursion for number of independent sets splits in two cases, with or without the root node being in the set.

a(10) has 113 digits and is too large to include.

Examples

			a(2) = a(1)^2 + a(0)^4 = 1^2 + 1^4 = 2.
a(3) = a(2)^2 + a(1)^4 = 2^2 + 1^4 = 5.
a(4) = a(3)^2 + a(2)^4 = 5^2 + 2^4 = 41.
a(5) = a(4)^2 + a(3)^4 = 41^2 + 5^4 = 2306.
a(6) = a(5)^2 + a(4)^4 = 2306^2 + 41^4 = 8143397.
a(7) = a(6)^2 + a(5)^4 = 8143397^2 + 2306^4 = 94592167328105.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..13
• Karl Petersen, Ibrahim Salama, Tree shift complexity, arXiv:1712.02251 [math.DS], 2017.
• Index entries for sequences of form a(n+1)=a(n)^2 + ...

Crossrefs

Cf. A000283, A005154, A112969, A114793.

Programs

Maple

A[0]:= 1: A[1]:= 1:
for n from 2 to 10 do
  A[n]:= A[n-1]^2 + A[n-2]^4;
od:
seq(A[i],i=0..10); # Robert Israel, Aug 21 2017

Mathematica

RecurrenceTable[{a[n] == a[n-1]^2 + a[n-2]^4, a[0] ==1, a[1] == 1}, a, {n, 0, 10}] (* Vaclav Kotesovec, Dec 18 2014 *)
NestList[{#[[2]],#[[1]]^4+#[[2]]^2}&,{1,1},10][[All,1]] (* Harvey P. Dale, Jul 03 2021 *)

PARI

{a(n) = if( n<2, 1, a(n-1)^2 + a(n-2)^4)}

PARI

{a=[0,0];for(n=1,99,iferr(a=[a[2],log(exp(a*[4,0;0,2])*[1,1]~)],E,return([n,exp(a[2]/2^n)])))} \\ To compute an approximation of the constant c1 = exp(lim_{n->oo} (log a(n))/2^n). \\ M. F. Hasler, May 21 2017

PARI

a=vector(20); a[1]=1;a[2]=2; for(n=3, #a, a[n]=a[n-1]^2+a[n-2]^4); concat(1, a) \\ Altug Alkan, Apr 04 2018

Formula

If b(n) = 1 + 1/b(n-1)^2, b(1)=1, then b(n) = a(n)/a(n-1)^2.

Lim_{n->inf} a(n)/a(n-1)^2 = A092526 (constant).

a(n) is asymptotic to c1^(2^n) * c2.

c1 = 1.2897512927198122075..., c2 = 1/A092526 = A263719 = (1/6)*(108 + 12*sqrt(93))^(1/3) - 2/(108 + 12*sqrt(93))^(1/3) = 0.682327803828019327369483739711... is the root of the equation c2*(1 + c2^2) = 1. - Vaclav Kotesovec, Dec 18 2014

Extensions

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 15 2007